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Question Number 143570 by cesarL last updated on 15/Jun/21

Answered by Ar Brandon last updated on 15/Jun/21

$$\mathrm{I}=\int {\tan }^2{8\mathrm{xsec}}^{4}8\mathrm{xdx}$$$$=\int {\tan }^2{8\mathrm{xsec}}^{2}8\mathrm{x}\cdot {\sec }^{2}8\mathrm{xdx}$$$$=\frac{1}{8}\int {\tan }^{2}8\mathrm{x}(1+{\tan }^{2}8\mathrm{x})\left({8\sec }^{2}8\mathrm{x}\right)\mathrm{dx}$$$$=\frac{1}{8}\int ({\tan }^{2}8\mathrm{x}+{\tan }^{4}8\mathrm{x})\mathrm{d}\left(\tan 8\mathrm{x}\right)$$$$=\frac{{\tan }^{3}8\mathrm{x}}{24}+\frac{{\tan }^{5}8\mathrm{x}}{40}+\mathrm{C}$$

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